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Jun 3, 2024
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feat: update solutions to lc problem: No.367 #3002

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108 changes: 44 additions & 64 deletions solution/0300-0399/0367.Valid Perfect Square/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -57,9 +57,9 @@ tags:

### 方法一：二分查找

不断循环二分枚举数字，判断该数的平方与 `num` 的大小关系，进而缩短空间，继续循环直至 $left \lt right$ 不成立。循环结束判断 $left^2$ 与 `num` 是否相等。
我们可以使用二分查找来解决这个问题。定义二分查找的左边界 $l = 1$，右边界 $r = num$，然后在 $[l, r]$ 的范围内查找满足 $x^2 \geq num$ 的最小整数 $x$。最后，如果 $x^2 = num$，则说明 $num$ 是一个完全平方数。

时间复杂度：$O(logN)$。
时间复杂度 $O(\log n)$，其中 $n$ 是给定的数字。空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand All @@ -68,31 +68,25 @@ tags:
```python
class Solution:
def isPerfectSquare(self, num: int) -> bool:
left, right = 1, num
while left < right:
mid = (left + right) >> 1
if mid * mid >= num:
right = mid
else:
left = mid + 1
return left * left == num
l = bisect_left(range(1, num + 1), num, key=lambda x: x * x) + 1
return l * l == num
```

#### Java

```java
class Solution {
public boolean isPerfectSquare(int num) {
long left = 1, right = num;
while (left < right) {
long mid = (left + right) >>> 1;
if (mid * mid >= num) {
right = mid;
int l = 1, r = num;
while (l < r) {
int mid = (l + r) >>> 1;
if (1L * mid * mid >= num) {
r = mid;
} else {
left = mid + 1;
l = mid + 1;
}
}
return left * left == num;
return l * l == num;
}
}
```
Expand All @@ -103,15 +97,16 @@ class Solution {
class Solution {
public:
bool isPerfectSquare(int num) {
long left = 1, right = num;
while (left < right) {
long mid = left + right >> 1;
if (mid * mid >= num)
right = mid;
else
left = mid + 1;
int l = 1, r = num;
while (l < r) {
int mid = l + (r - l) / 2;
if (1LL * mid * mid >= num) {
r = mid;
} else {
l = mid + 1;
}
}
return left * left == num;
return 1LL * l * l == num;
}
};
```
Expand All @@ -120,61 +115,44 @@ public:

```go
func isPerfectSquare(num int) bool {
left, right := 1, num
for left < right {
mid := (left + right) >> 1
if mid*mid >= num {
right = mid
} else {
left = mid + 1
}
}
return left*left == num
l := sort.Search(num, func(i int) bool { return i*i >= num })
return l*l == num
}
```

#### TypeScript

```ts
function isPerfectSquare(num: number): boolean {
let left = 1;
let right = num >> 1;
while (left < right) {
const mid = (left + right) >>> 1;
if (mid * mid < num) {
left = mid + 1;
let [l, r] = [1, num];
while (l < r) {
const mid = (l + r) >> 1;
if (mid >= num / mid) {
r = mid;
} else {
right = mid;
l = mid + 1;
}
}
return left * left === num;
return l * l === num;
}
```

#### Rust

```rust
use std::cmp::Ordering;
impl Solution {
pub fn is_perfect_square(num: i32) -> bool {
let num: i64 = num as i64;
let mut left = 1;
let mut right = num >> 1;
while left < right {
let mid = left + (right - left) / 2;
match (mid * mid).cmp(&num) {
Ordering::Less => {
left = mid + 1;
}
Ordering::Greater => {
right = mid - 1;
}
Ordering::Equal => {
return true;
}
let mut l = 1;
let mut r = num as i64;
while l < r {
let mid = (l + r) / 2;
if mid * mid >= (num as i64) {
r = mid;
} else {
l = mid + 1;
}
}
left * left == num
l * l == (num as i64)
}
}
```
Expand All @@ -185,11 +163,11 @@ impl Solution {

<!-- solution:start -->

### 方法二：转换为数学问题
### 方法二：数学

由于 `n² = 1 + 3 + 5 + ... + (2n-1)`，对数字 `num` 不断减去 $i$ (`i = 1, 3, 5, ...`) 直至 `num` 不大于 0，如果最终 `num` 等于 0，说明是一个有效的完全平方数。
由于 $1 + 3 + 5 + \cdots + (2n - 1) = n^2$，我们可以将 $num$ 逐渐减去 $1, 3, 5, \cdots$，如果最后 $num = 0$，则说明 $num$ 是一个完全平方数。

时间复杂度：$O(sqrt(N))$。
时间复杂度 $O(\sqrt n)$，空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand Down Expand Up @@ -224,7 +202,9 @@ class Solution {
class Solution {
public:
bool isPerfectSquare(int num) {
for (int i = 1; num > 0; i += 2) num -= i;
for (int i = 1; num > 0; i += 2) {
num -= i;
}
return num == 0;
}
};
Expand Down
119 changes: 47 additions & 72 deletions solution/0300-0399/0367.Valid Perfect Square/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -53,7 +53,11 @@ tags:

<!-- solution:start -->

### Solution 1: Binary search
### Solution 1: Binary Search

We can use binary search to solve this problem. Define the left boundary $l = 1$ and the right boundary $r = num$ of the binary search, then find the smallest integer $x$ that satisfies $x^2 \geq num$ in the range $[l, r]$. Finally, if $x^2 = num$, then $num$ is a perfect square.

The time complexity is $O(\log n)$, where $n$ is the given number. The space complexity is $O(1)$.

<!-- tabs:start -->

Expand All @@ -62,31 +66,25 @@ tags:
```python
class Solution:
def isPerfectSquare(self, num: int) -> bool:
left, right = 1, num
while left < right:
mid = (left + right) >> 1
if mid * mid >= num:
right = mid
else:
left = mid + 1
return left * left == num
l = bisect_left(range(1, num + 1), num, key=lambda x: x * x) + 1
return l * l == num
```

#### Java

```java
class Solution {
public boolean isPerfectSquare(int num) {
long left = 1, right = num;
while (left < right) {
long mid = (left + right) >>> 1;
if (mid * mid >= num) {
right = mid;
int l = 1, r = num;
while (l < r) {
int mid = (l + r) >>> 1;
if (1L * mid * mid >= num) {
r = mid;
} else {
left = mid + 1;
l = mid + 1;
}
}
return left * left == num;
return l * l == num;
}
}
```
Expand All @@ -97,15 +95,16 @@ class Solution {
class Solution {
public:
bool isPerfectSquare(int num) {
long left = 1, right = num;
while (left < right) {
long mid = left + right >> 1;
if (mid * mid >= num)
right = mid;
else
left = mid + 1;
int l = 1, r = num;
while (l < r) {
int mid = l + (r - l) / 2;
if (1LL * mid * mid >= num) {
r = mid;
} else {
l = mid + 1;
}
}
return left * left == num;
return 1LL * l * l == num;
}
};
```
Expand All @@ -114,61 +113,44 @@ public:

```go
func isPerfectSquare(num int) bool {
left, right := 1, num
for left < right {
mid := (left + right) >> 1
if mid*mid >= num {
right = mid
} else {
left = mid + 1
}
}
return left*left == num
l := sort.Search(num, func(i int) bool { return i*i >= num })
return l*l == num
}
```

#### TypeScript

```ts
function isPerfectSquare(num: number): boolean {
let left = 1;
let right = num >> 1;
while (left < right) {
const mid = (left + right) >>> 1;
if (mid * mid < num) {
left = mid + 1;
let [l, r] = [1, num];
while (l < r) {
const mid = (l + r) >> 1;
if (mid >= num / mid) {
r = mid;
} else {
right = mid;
l = mid + 1;
}
}
return left * left === num;
return l * l === num;
}
```

#### Rust

```rust
use std::cmp::Ordering;
impl Solution {
pub fn is_perfect_square(num: i32) -> bool {
let num: i64 = num as i64;
let mut left = 1;
let mut right = num >> 1;
while left < right {
let mid = left + (right - left) / 2;
match (mid * mid).cmp(&num) {
Ordering::Less => {
left = mid + 1;
}
Ordering::Greater => {
right = mid - 1;
}
Ordering::Equal => {
return true;
}
let mut l = 1;
let mut r = num as i64;
while l < r {
let mid = (l + r) / 2;
if mid * mid >= (num as i64) {
r = mid;
} else {
l = mid + 1;
}
}
left * left == num
l * l == (num as i64)
}
}
```
Expand All @@ -179,20 +161,11 @@ impl Solution {

<!-- solution:start -->

### Solution 2: Math trick
### Solution 2: Mathematics

This is a math problem：
Since $1 + 3 + 5 + \cdots + (2n - 1) = n^2$, we can gradually subtract $1, 3, 5, \cdots$ from $num$. If $num$ finally equals $0$, then $num$ is a perfect square.

```bash
1 = 1
4 = 1 + 3
9 = 1 + 3 + 5
16 = 1 + 3 + 5 + 7
25 = 1 + 3 + 5 + 7 + 9
36 = 1 + 3 + 5 + 7 + 9 + 11
....
so 1+3+...+(2n-1) = (2n-1 + 1)n/2 = n²
```
The time complexity is $O(\sqrt n)$, and the space complexity is $O(1)$.

<!-- tabs:start -->

Expand Down Expand Up @@ -227,7 +200,9 @@ class Solution {
class Solution {
public:
bool isPerfectSquare(int num) {
for (int i = 1; num > 0; i += 2) num -= i;
for (int i = 1; num > 0; i += 2) {
num -= i;
}
return num == 0;
}
};
Expand Down
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